An SOS Equation

As no square is negative, we have

$(\frac{a}{2}-b)^2=\frac{a^2}{4}+b^2-ab \ge 0$

$(\frac{a}{2}-c)^2=\frac{a^2}{4}+c^2-ac \ge 0$

$(\frac{d}{2}-c)^2=\frac{a^2}{4}+d^2-ad \ge 0$

adding the $3$ above equations, we have $\frac{3a^2}{4}+b^2+c^2+d^2-ab-ac-ad \ge 0$

And using the fact that $\frac{a^2}{4} \ge 0$ , we have $a^2+b^2+c^2+d^2-ab-ac-ad \ge 0$

Which yields, $a^2+b^2+c^2+d^2 \ge a(b+c+d)$ with equality if and only if $a=b=c=d=0$ .
But this is our given equation. Hence, the only solution is $(a, b,c,d)=(0,0,0,0)$ . So our answer is $\boxed{1}$